\documentclass[reqno]{amsart} \usepackage{amssymb} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2009(2009), No. 20, pp. 1--11.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2009 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2009/20\hfil An incorrectly posed problem] {An incorrectly posed problem for nonlinear elliptic equations} \author[S. G. Georgiev \hfil EJDE-2009/20\hfilneg] {Svetlin G. Georgiev} \address{Svetlin Georgiev Georgiev \newline Department of Differential Equations University of Sofia, Sofia, Bulgaria} \email{sgg2000bg@yahoo.com} \thanks{Submitted January 6, 2008. Published January 23, 2009.} \subjclass[2000]{35J60, 35J65, 35B05} \keywords{Nonlinear elliptic equation; incorrectly posed problems} \begin{abstract} We study properties of solutions to non-linear elliptic problems involving the Laplace operator on the unit sphere. In particular, we show that solutions do not depend continuously on the initial data. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \section{Introduction} In this paper we study properties of solutions to the initial-value problem \begin{gather} \label{e1.1} u_{rr}+{{n-1}\over r}u_r+{1\over {r^2}}\Delta_S u=f(r, u), \quad r\geq r_0,\\ u|_{r= r_0}=u_0\in X,\quad {u_r}|_{r= r_0}=u_1\in Y, \label{e1.2} \end{gather} where $n\geq 2$, $r_0\geq 1$ is suitable chosen and fixed number, $X$ and $Y$ are Banach spaces, $f\in \mathcal{C}([r_0, \infty))\times \mathcal{C}^1(\mathbb{R}^1)$, $f(r, 0)=0$ for every $r\geq r_0$, $a|u|\leq f_u'(r, u)\leq b|u|$ for every $r\geq r_0$, $u\in \mathbb{R}^1$, $a$ and $b$ are positive constants, $\Delta_S$ is the Laplace operator on the unit sphere $S^{n-1}$. More precisely we prove that the initial-value problem \eqref{e1.1}-\eqref{e1.2} is incorrectly posed in the following sense. When we say that \eqref{e1.1}-\eqref{e1.2} is incorrectly posed when the following happens: \eqref{e1.1}-\eqref{e1.2} has exactly one solution $u(r)\in X$ for each $u_0\in X$, $u_1\in Y$; there exists $\epsilon>0$ such that for every $\delta>0$, we have: $\|u_0-u_0'\|_X<\delta$, $\|u_1-u_1'\|_Y<\delta$ and $\|u-u'\|_{X}\geq \epsilon$, where $u$ is a solution with initial data $u_0, u_1$, and $u'$ is a solution with initial data $u_0', u_1'$. In this article, we obtain the following results using the same approach as in \cite{g1,g2,g3,g4}, \begin{theorem} \label{thm1.1} Let $n\geq 2$, $r_0\geq 1$, $f\in \mathcal{C}([r_0, \infty))\times \mathcal{C}^1(\mathbb{R}^1)$ $f(r, 0)=0$ for every $r\geq r_0$, and $X=Y=L^2(S^{n-1})$. Assume that there are positive constants, $a\leq b$, such that $a|u|\leq f_u'(r, u)\leq b|u|$ for every $r\geq r_0$ and every $u\in \mathbb{R}$. Then \eqref{e1.1}-\eqref{e1.2} is incorrectly posed. \end{theorem} \begin{theorem} \label{thm1.2} Let $n\geq 2$, $r_0\geq 1$, $f\in \mathcal{C}([r_0, \infty))\times \mathcal{C}^1(\mathbb{R}^1)$, $f(r, 0)=0$ for every $r\geq r_0$, $X=\mathcal{C}^2(S^{n-1})$ and $Y=\mathcal{C}^1(S^{n-1})$. Assume that there are positive constants, $a\leq b$, such that $a|u|\leq f_u'(r, u)\leq b|u|$ for every $r\geq r_0$ and every $u\in \mathbb{R}$. Then \eqref{e1.1}-\eqref{e1.2} is incorrectly posed. \end{theorem} This paper is organized as follows. In section 2 we prove our main results. In the appendix we prove results needed for the proof of Theorems \ref{thm1.1} and \ref{thm1.2}. \section{Proof of Main Results} Here and bellow we will assume that $r_0\geq 1$ and $n\geq 2$. First we will consider the initial-value problem \begin{gather} \label{e2.1} u_{rr}+{{n-1}\over r}u_r+{1\over {r^2}}\Delta_S u=f(u), \quad r\geq r_0,\\ \label{e2.2} u(r)|_{r= r_0}=u_0\in L^2(S^{n-1}), u_r(r)|_{r= r_0}=u_1\in {L}^{2}(S^{n-1}), \end{gather} where $\Delta_S$ is the Laplace operator on the unit sphere $S^{n-1}$, $f\in \mathcal{C}^1(\mathbb{R}^1)$, $f(0)=0$, $a|u|\leq f'(u)\leq b|u|$ for every $u\in \mathbb{R}^1$, $a\leq b$ are fixed positive constants. For fixed positive constants $n\geq 2$, $r_0\geq 1$, $a$, $b$, $a\leq b$, we suppose that the constants $A$, $B$, $c_1$, $d_1$ satisfy the following conditions $$\begin{gathered} r_0\leq c_1\leq d_1,\\ A\geq B>0,\\ {a\over {2A}}{{d_1^n}\over {(d_1+1)^n}}\geq 1. \end{gathered} \label{ei1}$$ \noindent\textbf{Example.} Let $n\geq 1$, $r_0\gg 1$, $A=2$, $B=1$, $a=r_0^{10n}$,$b=2r_0^{10n}$, $c_1=r_0+1$, $d_1=r_0+2$. Let $N$ be the set \begin{align*} N=\Bigl\{&u(r): u(r)\in \mathcal{C}^{2}([r_0, \infty)),\; u(\infty)=u_r(\infty)=0, \\ & r^{\alpha}|\partial_r^{\beta}u(r)|\leq 1 \; \forall r\geq r_0,\; \forall \alpha \in \mathbb{N}\cup \{0\},\beta=0, 1,\\ &u(r)\geq 0\; \forall r\geq r_0, u(r)\leq {1\over B} \; \forall r\geq r_0,\\ &u(r)\geq {1\over A}\; \forall r\in [c_1, d_1],\; u(r)\in L^2([r_0, \infty))\; \Bigr\}. \end{align*} For $n\geq 1$, $f(u)\in \mathcal{C}^1(\mathbb{R}^1)$, $a|u|\leq f'(u)\leq b|u|$, where $a\geq b$ are positive constants, and $u\in N$ we define the operator and the initial values \begin{gather*} P(u)=\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}\tau^n f(u)d\tau ds,\\ u_0=\int_{r_0}^{\infty}{1\over {s^n}}\int_s^{\infty}\tau^n f(u)d\tau ds,\quad u_1=-{1\over {r_0^n}}\int_{r_0}^{\infty}\tau^n f(u)d\tau. \end{gather*} \begin{theorem} \label{thm2.1} Let $n\geq 2$, $r_0\geq 1$, $f\in \mathcal{C}^1(\mathbb{R}^1)$, and $f(0)=0$. Assume that there exist positive constants $a\leq b$ such that $a|u|\leq f'(u)\leq b|u|$. Then \eqref{e2.1}-\eqref{e2.2} has exactly one solution $u\in N$. \end{theorem} \begin{proof} First we prove that $P:N\to N$. Let $u\in N$ be fixed. Then \\ \textbf{(1)} Since $f\in \mathcal{C}^1([r_0, \infty))$, $u\in \mathcal{C}^2([r_0, \infty))$ we have that $P(u)\in \mathcal{C}^2([r_0, \infty))$. Also we have \begin{gather*} P(u)_{|_{r=\infty}}=0,\\ {{\partial P(u)}\over {\partial r}}=-{1\over {r^n}}\int_r^{\infty} \tau^n f(u)d\tau,\\ {{\partial P(u)}\over {\partial r}}_{|_{r=\infty}}=0. \end{gather*} \noindent\textbf{(2)} Let $\alpha\in \mathbb{N}\cup\{0\}$. We choose $k\in \mathbb{N}$ such that $k\geq \alpha+3$ and ${b\over {2B(k-1)}}<1$. Then $$r^{\alpha}P(u)=r^{\alpha}\int_r^{\infty}{1\over {s^n}}\int_s^{\infty} \tau^n f(u)d\tau ds\,.$$ Now we use that for $u\in N$, we have $u\geq 0$ for every $r\geq r_0$, $f(0)=0$, $f'(u)\leq bu$, from here $f(u)\leq {b\over 2}u^2$, since $u\leq {1\over B}$ for every $r\geq r_0$ we get $f(u)\leq {b\over {2B}}u$. Then \begin{align*} r^{\alpha}P(u) &\leq {b\over {2B}}r^{\alpha}\int_r^{\infty}{1\over {s^n}} \int_s^{\infty}\tau^n ud\tau ds \\ &=r^{\alpha}{b\over {2B}}\int_r^{\infty}{1\over {s^n}}\int_s^{\infty} \tau^{n+k}{1\over {\tau^k}} ud\tau ds \quad \text{(use that $\tau^{n+k}u\leq 1$)}\\ &\leq {b\over {2B}}r^{\alpha}\int_r^{\infty}{1\over {s^n}} \int_s^{\infty}{1\over {\tau^k}} d\tau ds \\ & \leq {b\over {2B}}{1\over {(k-1)(n+k-2)}}{1\over {r_0^{n+k-\alpha-2}}} \leq 1. \end{align*} In the above inequality we use our choice of the constant $k$. Also, \begin{align*} \big|r^{\alpha}{{\partial P(u)}\over {\partial r}}\big| &\leq {b\over {2B}}r^{\alpha}{1\over {r^n}}\int_r^{\infty}\tau^n ud\tau \\ &=r^{\alpha}{b\over {2B}}{1\over {r^n}}\int_r^{\infty}\tau^{n+k} {1\over {\tau^k}} ud\tau \quad \text{(use $\tau^{n+k}u\leq 1$)}\\ & \leq r^{\alpha}{b\over {2B}}\int_r^{\infty}{1\over {s^n}} \int_s^{\infty}{1\over {\tau^k}} d\tau ds \\ & \leq {b\over {2B}}{1\over {(k-1)}}{1\over {r_0^{n+k-\alpha-1}}}\leq 1. \end{align*} In the above inequality we use our choice of the constant $k$. \noindent\textbf{(3)} First we note that for $u\in N$ we have $f(u)\geq a u^2/2$. Therefore for every $r\geq r_0$ we have $$P(u)\geq {a\over 2}\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}\tau^n u^2 d\tau ds\geq 0.$$ \noindent\textbf{(4)} Let $r\in [c_1, d_1]$. Then $$P'(u)=\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}\tau^n f'(u)d\tau ds\geq a\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}\tau^n ud\tau ds\geq 0.$$ Therefore, for $u\in N$ the function $P(u)$ is increase function of $u$. Since for every $r\in [c_1, d_1]$ we have that $u\geq 1/A$ we get \begin{align*} P(u)&\geq P({1\over A}) =\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}\tau^n f \bigl({1\over A}\bigr)d\tau ds \\ &\geq {a\over {2A^2}}\int_{d_1}^{d_1+1}{1\over {s^n}} \int_{d_1}^{d_1+1}\tau^n d\tau ds \\ & \geq {a\over {2A^2}}{{d_1^n}\over {(d_1+1)^n}}\geq {1\over A}, \end{align*} in the above inequality we use \eqref{ei1}. \noindent\textbf{(5)} Choose $k\in \mathbb{N}$ such that $$k>3,\quad {b\over {2(k-1)(n+k-2)}}<1.$$ Then \begin{align*} P(u)&\leq {b\over {2B}}\int_r^{\infty}{1\over {s^n}}\int_s^{\infty} \tau^n ud\tau ds \\ &\leq {b\over {2B}}\int_r^{\infty}{1\over {s^n}}\int_s^{\infty} \tau^{k+n}{1\over {\tau ^k}} ud\tau ds \\ &\leq {b\over {2B}}\int_r^{\infty}{1\over {s^n}}\int_s^{\infty} {1\over {\tau ^k}}d\tau ds \\ &={b\over {2B(k-1)(n+k-2)r_0^{n+k-2}}}\leq {1\over B}. \end{align*} \noindent\textbf{(6)} Now we prove that $P(u)\in L^2([r_0, \infty))$. Indeed, \begin{align*} \|P(u)\|_{L^2([r_0, \infty))}^2 &=\int_{r_0}^\infty \Bigl(\int_r^{\infty}{1\over {s^n}} \int_s^{\infty}\tau^n f(u)d\tau ds\Bigr)^2 dr \\ & \leq {{b^2}\over 4} \int_{r_0}^\infty \Bigl(\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}\tau^n u^2d\tau ds \Bigr)^2 dr \\ & \leq {{b^2}\over 4} \int_{r_0}^\infty \Bigl(\int_r^{\infty}{1\over {s^n}} \int_s^{\infty}\tau^{k+n}u {u\over {\tau^k}}d\tau ds\Bigr)^2 dr\quad \text{(use that $\tau^{k+n}u\leq 1$)}\\ &\leq {{b^2}\over 4} \int_{r_0}^\infty \Bigl(\int_r^{\infty}{1\over {s^n}} \int_s^{\infty} {u\over {\tau^k}}d\tau ds\Bigr)^2 dr\leq \quad \text{(use H\"older's inequality)}\\ & \leq {{b^2}\over 4} \int_{r_0}^\infty \Bigl(\int_r^{\infty}{1\over {s^n}} \Bigl(\int_s^{\infty}{1\over {\tau^{2k}}}d\tau\Bigr)^{1/2} \Bigl(\int_s^{\infty}u^2d\tau\Bigr)^{1/2} ds\Bigr)^2 dr \\ & \leq {{b^2}\over {4(2k-1)(n+k-{3\over 2})^2(2n+2k-4)r_0^{2n+2k-4}}} \|u\|_{L^2([r_0, \infty))}^2<\infty, \end{align*} because $u\in L^2([r_0, \infty))$. From (1)--(6) we conclude that $P:N\to N$. Now we prove that the operator $P$ has exactly one fixed point in $N$. Let $u_1, u_2\in N$ are fixed and $\alpha=\|u_1-u_2\|_{L^2([r_0, \infty))}$. We choose the constant $k\in \mathbb{N}$ large so that $Q_1/\alpha<1$, where $$Q_1={{2b^2}\over {B^2({4\over 3}k-1)^{3\over 2}(n+k-{7\over 4})^2(2n+2k-{9\over 2}) r_0^{2n+2k-{9\over 2}}}}.$$ Then \begin{align*} &\|P(u_1)-P(u_2)\|_{L^2([r_0, \infty))}^2 \\ &=\int_{r_0}^{\infty}\Bigl(\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}\tau^n (f(u_1)-f(u_2))d\tau ds\Bigr)^2 dr\quad \text{(mean value theorem)}\\ &=\int_{r_0}^{\infty}\Bigl(\int_r^{\infty}{1\over {s^n}} \int_s^{\infty}\tau^n f'(\xi)(u_1-u_2)d\tau ds\Bigr)^2 dr \\ & \leq \int_{r_0}^{\infty}\Bigl(\int_r^{\infty}{1\over {s^n}} \int_s^{\infty}\tau^n |f'(\xi)\|u_1-u_2|d\tau ds\Bigr)^2 dr \\ &\quad \text{(use that $|f'(\xi)|\leq b|\xi|\leq {b\over B}$, $|\xi|\leq \max\{|u_1|, |u_2|\}$)}\\ &\leq {{b^2}\over {B^2}} \int_{r_0}^{\infty}\Bigl(\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}\tau^n |u_1-u_2|d\tau ds\Bigr)^2 dr \\ &={{b^2}\over {B^2}} \int_{r_0}^{\infty}\Bigl(\int_r^{\infty}{1\over {s^n}}\int_s^{\infty} \sqrt{\tau^{2k+2n} |u_1-u_2|}{1\over {\tau^k}}\sqrt{|u_1-u_2|}d\tau ds\Bigr)^2 dr \\ &\quad \text{(use that $\sqrt{\tau^{2k+2n}|u_1-u_2|}\leq \sqrt{2}$)}\\ &\leq {{2b^2}\over {B^2}} \int_{r_0}^{\infty}\Bigl(\int_r^{\infty}{1\over {s^n}}\int_s^{\infty} \sqrt{|u_1-u_2|}{1\over {\tau^k}}d\tau ds\Bigr)^2 dr \quad \text{(H\"older's inequality)}\\ &\leq {{2b^2}\over {B^2}} \int_{r_0}^{\infty}\Bigl(\int_r^{\infty}{1\over {s^n}}\Bigl(\int_s^{\infty} {1\over {\tau^{{4k}\over 3}}}d\tau\Bigr)^{3/4} \Bigl({|u_1-u_2|}^2d\tau\Bigr)^{1/4} ds\Bigr)^2 dr \\ & \leq Q_1\|u_1-u_2\|_{L^2([r_0, \infty))}; \end{align*} i.e., $$\|P(u_1)-P(u_2)\|_{L^2([r_0, \infty))}^2\leq Q_1\|u_1-u_2\|_{L^2([r_0, \infty))}.$$ From this, $$\|P(u_1)-P(u_2)\|_{L^2([r_0, \infty))}^2\leq {{Q_1}\over {\alpha}}\alpha\|u_1-u_2\|_{L^2([r_0, \infty))}\leq {{Q_1}\over {\alpha}}\|u_1-u_2\|_{L^2([r_0, \infty))}^2.$$ For our next step we need the theorem in \cite[page 294]{k1}: \begin{quote} Let $B$ be the complete metric space for which $AB\subset B$ and for the operator $A$ satisfies the condition $$\rho(Ax, Ay)\leq L(\alpha, \beta)\rho(x, y),\quad x, y\in B,\alpha\leq \rho(x, y)\leq \beta,$$ where $L(\alpha, \beta)<1$ for $0<\alpha\leq\beta<\infty$. Then the operator $A$ has exactly one fixed point in $B$. \end{quote} From the above result and our choice of $k$ we conclude that the operator $P$ has exactly one fixed point $u\in N$. Consequently $u$ is a solution to the problem \eqref{e2.1}-\eqref{e2.2}. In the appendix we will prove that the set $N$ is closed subset of the space $L^2([r_0, \infty))$. We have that $u_0\in L^2(S^{n-1})$, $u_1\in L^2(S^{n-1})$. \end{proof} \begin{theorem} \label{thm2.2} Let $n\geq 2$, $r_0\geq 1$, $f\in \mathcal{C}^1(\mathbb{R}^1)$, and $f(0)=0$. Assume that there exists positive constants, $a\leq b$, such that $a|u|\leq f'(u)\leq b|u|$. Then \eqref{e2.1}-\eqref{e2.2} is incorrectly posed. \end{theorem} \begin{proof} On the contrary, suppose that \eqref{e2.1}-\eqref{e2.2} is correctly posed. Let $u$ is the solution from Theorem \ref{thm2.1}. We choose $\epsilon$ such that $0<\epsilon<1/Q_2$, where $$Q_2={{b^2}\over {4(4k-1)^{1/2}(n+k-{5\over 4})^2 (2n+2k-{7\over 2})r_0^{2n+2k-{7\over 2}}}}.$$ Then there exists $\delta=\delta(\epsilon)>0$ such that $$\|u_0\|_{L^2(S^{n-1})}<\delta, \quad \|u_1\|_{L^2(S^{n-1})}<\delta$$ imply $$\|u\|_{L^2([r_0, \infty))}<\epsilon.$$ From the definition of $u$, we have \begin{align*} \|u\|_{L^2([r_0, \infty))}^2 &=\int_{r_0}^{\infty} \Bigl(\int_r^{\infty}{1\over {s^n}} \int_s^{\infty}\tau^n f(u)d\tau ds\Bigr)^2 dr \\ &\leq {{b^2}\over 4} \int_{r_0}^{\infty} \Bigl(\int_r^{\infty}{1\over {s^n}}\int_s^{\infty} \tau^n u^2d\tau ds\Bigr)^2 dr \\ &= {{b^2}\over 4} \int_{r_0}^{\infty} \Bigl(\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}\sqrt{\tau^{2k+2n}u} u^{3\over 2}{1\over {\tau^k}}d\tau ds\Bigr)^2 dr \\ &\quad \text{(use that $\sqrt{\tau^{2k+2n}u} \leq 1$)}\\ &\leq {{b^2}\over 4} \int_{r_0}^{\infty} \Bigl(\int_r^{\infty}{1\over {s^n}}\int_s^{\infty} u^{3\over 2}{1\over {\tau^k}}d\tau ds\Bigr)^2 dr \quad \text{(H\"older's inequality)}\\ &\leq {{b^2}\over 4} \int_{r_0}^{\infty} \Bigl(\int_r^{\infty}{1\over {s^n}} \Bigl(\int_s^{\infty} u^{2}d\tau\Bigr)^{3/4} \Bigl({1\over {\tau^{4k}}}d\tau\Bigr)^{1/4} ds\Bigr)^2 dr \\ &\leq Q_2\|u\|_{L^2([r_0, \infty))}^3; \end{align*} i.e., $$\|u\|_{L^2([r_0, \infty))}^2\leq {{Q_2}}\|u\|_{L^2([r_0, \infty))}^3.$$ From this,, $$\|u\|_{L^2([r_0, \infty))}\geq {1\over {Q_2}}>\epsilon$$ which is a contradiction. Consequently the problem \eqref{e2.1}-\eqref{e2.2} is incorrectly posed. \end{proof} \begin{theorem} \label{thm2.3} Let $n\geq 2$, $r_0\geq 1$, $f\in \mathcal{C}^1(\mathbb{R}^1)$, and $f(0)=0$. Assume that there are positive constants $a\leq b$ such that $a|u|\leq f'(u)\leq b|u|$. Then the problem \begin{gather} u_{rr}+{{n-1}\over r}u_r+{1\over {r^2}}\Delta_S u=f(u), \quad r\geq r_0, \label{e2.1b} \\ \label{e2.3} u(r)_{|_{r= r_0}}=u_0\in \mathcal{C}^2(S^{n-1}),\quad u_r(r)_{|_{r=r_0}}=u_1\in \mathcal{C}^{1}(S^{n-1}), \end{gather} is incorrectly posed. \end{theorem} \begin{proof} Let us suppose that \eqref{e2.1b}-\eqref{e2.3} is correctly posed, and let $$Q_3={b\over {2(k-1)(n+k-2)r_0^{n+k-2}}}.$$ Then for $0<\epsilon<1/Q_3^2$, there exists $\delta=\delta(\epsilon)>0$ such that $$\|u_0\|_{\mathcal{C}^2(S^{n-1})}<\delta,\quad \|u_1\|_{\mathcal{C}^1(S^{n-1})}<\delta,$$ imply $$\max_{r\in [r_0, \infty)}|u|<\epsilon,\quad \max_{r\in [r_0, \infty)}|u_r|<\epsilon,\quad \max_{r\in [r_0, \infty)}|u_{rr}|<\epsilon,$$ where $u$ is the solution from the Theorem \ref{thm2.1}. From the definition of $u$, and $k\in \mathbb{N}$, we have \begin{align*} u(r)&\leq {b\over 2}\int_r^{\infty}{1\over {s^n}} \int_s^{\infty}\tau^n u^2 d\tau ds\\ &={b\over 2}\int_r^{\infty}{1\over {s^n}} \int_s^{\infty}\sqrt {\tau^{2k+2n}u} u^{3\over 2}{1\over {\tau ^k}} d\tau ds \\ & \leq {b\over 2}\int_r^{\infty}{1\over {s^n}}\int_s^{\infty} u^{3\over 2}{1\over {\tau^k}} d\tau ds\\ &\leq {b\over 2}(\max_{r\in [r_0, \infty)}u)^{3\over 2} \int_r^{\infty}{1\over {s^n}}\int_s^{\infty} {1\over {\tau^k}} d\tau ds\\ &\leq Q_3(\max_{r\in [r_0, \infty)}u)^{3\over 2}. \end{align*} From this it follows that $$Q_3(\max_{r\in [r_0, \infty)}u)^{1/2}\geq 1,\quad \text{or}\quad \max_{r\in [r_0, \infty)}u>{1\over {Q_3^2}}>\epsilon,$$ which is a contradiction with our assumption. Consequently \eqref{e2.1b}-\eqref{e2.3} is incorrectly posed. \end{proof} The proofs of Theorems \ref{thm1.1} and \ref{thm1.2} follow from the method used in the proof of Theorems \ref{thm2.2} and \ref{thm2.3}. \section{Appendix} \begin{lemma} \label{lem3.1} The set $N$ is a closed subset of $L^2([r_0, \infty))$. \end{lemma} \begin{proof} Let $\{u_n\}$ is a sequence of elements in $N$ for which $$\lim_{n\to\infty}\|u_n-\tilde{u}\|_{L^2([r_0, \infty))}=0,$$ where $\tilde{u}\in L^2([r_0, \infty))$. Since $P(u)$ is a continuous differentiable function of $u$, for $r\in [r_0, c_1]$ and $u\in N$ we have \begin{align*} P'(u)&=\int_r^{\infty}{1\over {s^{n}}}\int_s^{\infty}\tau^{n} f'(u)d\tau ds \\ & \geq a\int_{c_1}^{d_1}{1\over {s^{n}}}\int_{c_1}^{d_1}\tau^{n} u d\tau ds \\ & \geq {a\over A}{{c_1^n}\over {d_1^n}}(d_1-c_1)^2. \end{align*} From this, it follows that for every $u\in N$ there exists $$L=\min_{r\in [r_0, c_1]}|P'(u)(r)|>0.$$ Let $$M_1=\max_{r\in [r_0, c_1]}\Bigl|{{\partial}\over {\partial r}}P'(u)(r)\Bigr|.$$ Now we prove that for every $\epsilon >0$ there exists $\delta=\delta(\epsilon)>0$ such that from $|x-y|<\delta$ we have $$|u_m(x)-u_m(y)|<\epsilon\quad \forall m\in \mathbb{N}.$$ We suppose that there exists ${\tilde \epsilon}>0$ such that for every $\delta>0$ there exist natural number $m$ and $x, y\in [r_0, \infty)$, $|x-y|<\delta$ for which $|u_m(x)-u_m(y)|\geq {\tilde\epsilon}$. We choose ${\tilde{\tilde \epsilon}}$ such that $0<{\tilde {\tilde \epsilon}}0$ such that for every natural $m$ we have $$|P(u_m)(x)-P(u_m)(y)|<{\tilde {\tilde \epsilon}},\quad \forall x, y\in [r_0, \infty):|x-y|<\delta_1.$$ Consequently we can choose $$0<\delta<\min\bigl\{c_1-r_0, \delta_1, {{(L{\tilde\epsilon}-{\tilde {\tilde\epsilon}})B} \over {M_1}}\bigr\}$$ such that there exist natural number $m$ and $x_1, x_2\in [r_0, \infty)$ for which $$|x_1-x_2|<\delta,\quad |u_m(x_1-x_2+r_0)-u_m(r_0)|\geq {\tilde\epsilon}.$$ In particular, $$|P(u_m)(x_1-x_2+r_0)-P(u_m)(r_0)|<{\tilde {\tilde \epsilon}}. \label{e3.1}$$ Let us suppose for convenience that $x_1-x_2>0$. Then $x_1-x_20$ there exists $\delta=\delta(\epsilon)>0$ such that from $|x-y|<\delta$ follows $$|u_m(x)-u_m(y)|<\epsilon\quad \forall m\in \mathbb{N}. \label{e3.2}$$ On the other hand from the definition of the set $N$ we have that for every natural number $m$ $$u_m(r)\leq {1\over B}\quad \forall r\geq r_0. \label{e3.3}$$ From this inequality and \eqref{e3.2} it follows that the set $\{u_m\}$ is a compact subset of the space $\mathcal{C}([r_0, \infty))$. Therefore there is a subsequence $\{u_{n_k}\}$ and function $u\in \mathcal{C}([r_0, \infty))$ for which $$|u_{n_k}(x)-u(x)|<\epsilon\quad \forall x\in [r_0, \infty).$$ Now we suppose that that $u\neq \tilde{u}$ a.e. in $[r_0, \infty)$. Then there exist $\epsilon_1>0$ and subinterval $\Delta\subset [r_0, \infty)$ such that $\mu(\Delta)>0$ and $$|u-\tilde{u}|>\epsilon_1\quad \text{for } r\in \Delta.$$ Let $\epsilon>0$ is chosen such that $$\epsilon<{{\epsilon_1(\mu(\Delta))^{1/2}}\over {\mu(\Delta)^{1/2}+1}}. \label{e3.4}$$ Then, for every $n_k\in \mathbb{N}$ sufficiently large, we have $\|u_{n_k}-\tilde{u}\|_{L^2([r_0, \infty))}<\epsilon$, \begin{align*} \epsilon\mu(\Delta) &=\epsilon\int_{\Delta} dx\\ &>\int_{\Delta}|u_{n_k}-{u}|dx =\int_{\Delta}|u_{n_k}-\tilde{u}+\tilde{u}-u|dx\\ &\geq \int_{\Delta}|\tilde{u}-u|dx-\int_{\Delta}|u_{n_k}-\tilde{u}|dx \\ &\geq \epsilon_1\mu(\Delta)-\Bigl(\int_{\Delta}|u_{n_k}-\tilde{u}|^2dx\Bigr) ^{1/2} \bigl(\mu(\Delta)\bigr)^{1/2} \\ &\geq\epsilon_1\mu(\Delta)-\|u_{n_k}-\tilde{u}\|_{L^2([r_0, \infty))} \bigl(\mu(\Delta)\bigr)^{1/2} \\ &>\epsilon_1\mu(\Delta)-\epsilon\bigl(\mu(\Delta)\bigr)^{1/2}, \end{align*} which is a contradiction with \eqref{e3.4}. From this, $u=\tilde{u}$ a.e. in $[r_0, \infty)$, $|u_n-u|^2=|\tilde{u}-u_n|^2$ a.e. in $[r_0, \infty)$, $\|u_n-u\|_{L^2([r_0,\infty))}=\|u_n-\tilde{u}\|_{L^2([r_0, \infty))}$. Consequently, for every sequence $\{u_n\}$ from elements of the set $N$, which is convergent in $L^2([r_0, \infty))$, there exists a function $u\in \mathcal{C}([r_0, \infty))$, $u\in L^2([r_0, \infty))$ for which $$\lim_{n\to \infty}\|u_n-u\|_{L^2([r_0, \infty))}=0.$$ Bellow we will suppose that $\{u_n\}$ is a sequence from elements of the set $N$, which is convergent in $L^2([r_0, \infty))$. Then there exists a function $u\in \mathcal{C}([r_0, \infty))$, $u\in L^2([r_0, \infty))$ for which $$\lim_{n\to \infty}\|u_n-u\|_{L^2([r_0, \infty))}=0.$$ Now we suppose that $u(\infty)\ne 0$. Then there exist sufficiently large $Q>0$, a large natural number $m$ and $\epsilon_2>0$ for which $$u_m(r)=0,\quad u(r)>\epsilon_2,\quad \forall r\geq Q.$$ We choose $$0<\epsilon_3<\epsilon_2. \label{e3.5}$$ Then, for every $n\in \mathbb{N}$ sufficiently large, we have $|u_n(r)-u( r)|<\epsilon_3$ and \begin{align*} \epsilon_3&>\int_Q^{Q+1}|u_n(r)-u(r)|dr \\ &\geq \int_Q^{Q+1}(|u(r)|-|u_n( r)|)dr \\ &=\int_Q^{Q+1}|u(r)|dr>\epsilon_2, \end{align*} which is a contradiction with \eqref{e3.5}. Therefore, $u(\infty)=0$. Now we prove that ${{\partial}\over {\partial r}}u(r)$ exists for every $r\geq r_0$. Let us suppose that there exists $r_1\in [r_0, \infty)$ such that ${{\partial}\over {\partial r}}u(r_1)$ does not exists. Then for every $h>0$, which is enough small, exists $\epsilon_4>0$ such that $$\Bigl|{{u(r_1+h)-u(r_1)}\over h}\Bigr|>\epsilon_4,$$ and $$0<\epsilon_5<{h\over 2}\epsilon_4, \label{e3.6}$$ such that $|{{u_n(r_1+h)-u(r_1)}}|<\epsilon_5$. From this, \begin{align*} \epsilon_5&>|u_n(r_1+h)-u(r_1+h)| \\ &=|u_n(r_1+h)-u(r_1)+u(r_1)-u(r_1+h)| \\ &\geq |u(r_1)-u(r_1+h)|{1\over h} h-|u_n(r_1+h)-u(r_1)|\\ &\geq \epsilon_4 h-\epsilon_5, \end{align*} which is a contradiction of our choice of $\epsilon_5$. Therefore ${{\partial}\over {\partial r}}u(r)$ exists for every $r\in [r_0, \infty)$. As in above we can see that $u(r)\in \mathcal{C}^2([r_0, \infty))$ $u_r(\infty)=0$. Now we suppose that there exists interval $\Delta_2\subset [r_0, \infty)$ such that $$u(r)\geq {1\over B}+\epsilon_7\quad \text{for }r\in \Delta_2.$$ Let $n\in \mathbb{N}$ be large and $\epsilon_8>0$ chosen such that $$|u_n(r)-u(r)|<\epsilon_8\quad \text{for } r\in \Delta_2, 0<\epsilon_8<\epsilon_7. \label{e3.7}$$ From this, for $r\in \Delta_2$, we have $$\epsilon_8>|u_n(r)-u(r)|\geq |u(r)|-|u_n(r)|\geq {1\over B}+\epsilon_7-{1\over B}=\epsilon_7,$$ which is a contradiction with \eqref{e3.7}. Therefore, $u(r)\leq {1\over B}$ for every $r\geq r_0$. Now we suppose that there exists interval $\Delta_3\subset [c_1, d_1]$ for which $u(r)< {1\over A}$ for every $r\in \Delta_3$. From this, there exists $\epsilon_9>0$ such that $u(r)\leq {1\over A}-\epsilon_9$ for $r\in \Delta_3$. Also, let $$0<\epsilon_{10}<\epsilon_9 \label{e3.8}$$ and $n\in \mathbb{N}$ is enough large such that $\epsilon_{10}>|u_n(r)-u(r)|$ for $r\in \Delta_3$. Then for $r\in \Delta_3$ we have $$\epsilon_{10}>|u_n(r)-u(r)|\geq |u_n(r)|-|u(r)|\geq {1\over A}-{1\over A}+\epsilon_9,$$ which is a contradiction with \eqref{e3.8}. Consequently, for every $r\in [c_1, d_1]$ we have $u(r)\geq {1\over A}$. Now we suppose that there exist $\alpha\in \mathbb{N}\cup\{0\}$, interval $\Delta_4\subset [r_0, \infty)$ and $\epsilon_{11}>0$ such that $$|r^{\alpha}u( r)|>1+\epsilon_{11}\quad\text{for } r\in \Delta_4.$$ Let $\epsilon_{12}>0$ and $n\in \mathbb{N}$ be chosen such that $$|r^{\alpha}(u_n(r)-u(r))|<\epsilon_{12}\quad\text{for } r\in\Delta_4,\; 0<\epsilon_{12}<\epsilon_{11}. \label{e3.9}$$ From this, $$\epsilon_{12}>|r^{\alpha}(u_n(r)-u(r))|\geq |r^{\alpha}u(r)|- r^{\alpha}|u_n(r)|\geq\epsilon_{11},$$ which is a contradiction with \eqref{e3.9}. Therefore for every $\alpha\in \mathbb{N}\cup\{0\}$ and for every $r\in [r_0, \infty)$ we have $r^{\alpha}u(r)\leq 1$. After we use the same arguments we can see that for every $\alpha\in \mathbb{N}\cup\{0\}$ and for every $r\in [r_0, \infty)$ we have $r^{\alpha}|u_r(r)|\leq 1$. Now we suppose that there exist interval $\Delta_5\subset [r_0, \infty)$ and $\epsilon_{13}>0$ such that for $r\in \Delta_5$ we have $u(r)<-\epsilon_{13}$. Let $n\in \mathbb{N}$ is enough large and $\epsilon_{14}>0$ are fixed for which $$|u_n(r)-u(r)|<\epsilon_{14}\quad \text{for } r\in \Delta_5,\quad 0<\epsilon_{14}<\epsilon_{13}. \label{e3.10}$$ Then for $r\in \Delta_5$ we have $$\epsilon_{14}>u_n(r)-u(r)>\epsilon_{13}$$ which is a contradiction with \eqref{e3.10}. \end{proof} \begin{thebibliography}{00} \bibitem{b1} Bandle, C., M. Essen. On the solutions of quasilinear elliptic problems with boundary blow - up, PDE of Elliptic Type, (Cortona, 1992), Sympos. Math., Vol. 35, Cambridge University Press, 1994, pp. 93-111. \bibitem{c1} Cirstea, F., V. Radulescu; Uniqueness of the blow - up boundary solution of logistic equations with absorption, C. R. Acad. Sci. Paris, Ser. I 335(2002), 447-452. \bibitem{g1} Georgiev, S. Blow up of solutions for Klein-Gordon equations in the Reissner-Nordstr\"om metric, Electron. J. Diff. 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